GPT 6 Astra Ultra attempt on Math-6080

Erdős Problem 1026

Statement and precise scope.

For a sequence of distinct real numbers, consider the largest sum of a nonempty increasing or decreasing subsequence. The original request to "determine" this quantity has more than one possible interpretation. We give an exact recurrence for any given sequence and prove the sharp universal asymptotic constant in the clarified question: \[ M(x_1,\ldots,x_n)\geq\frac1{\sqrt n}\sum_{i=1}^n x_i. \tag{1} \] For positive sequences of square length $n=k^2$ with total sum one, this proves the requested lower bound $1/k$; no constant larger than one can replace the coefficient in $1$ uniformly as $n\to\infty$. The sharper extremal constants for individual nonsquare lengths are separate from the assertion proved here.

The exact value for any input sequence.

Let $I_i$ and $D_i$ be the largest sums of increasing and decreasing subsequences ending at position $i$, respectively. Then \[ I_i=x_i+\max\left(\{0\}\cup\{I_j:j<i,\ x_j<x_i\}\right), \] \[ D_i=x_i+\max\left(\{0\}\cup\{D_j:j<i,\ x_j>x_i\}\right). \tag{2} \] Every such subsequence either consists of its final entry or extends one ending at an eligible earlier entry. Conversely each choice in $2$ produces such a subsequence. Induction on $i$ therefore proves the recurrence, including for negative entries. The desired value is exactly $M=\max_i\max(I_i,D_i)$. Computing all the maxima by scanning earlier positions takes $O(n^2)$ comparisons and additions; retaining a maximizing predecessor recovers an actual subsequence.

A stronger weighted inequality for positive entries.

Assume first that all $x_i>0$, and write $I=\max I_i$, $D=\max D_i$. Associate to position $i$ the square \[ Q_i=(I_i-x_i,I_i]\times(D_i-x_i,D_i]. \] It has area $x_i^2$, and lies in $(0,I]\times(0,D]$, since both recurrences include the option zero.

These squares are pairwise disjoint. Indeed, for $i<j$, if $x_i<x_j$, then $2$ gives $I_j-x_j\geq I_i$, so the two horizontal intervals are disjoint. If $x_i>x_j$, then $D_j-x_j\geq D_i$, so the two vertical intervals are disjoint. The half-open convention handles equality at a boundary without double counting.

Adding areas now gives \[ ID\geq\sum_i x_i^2, \qquad M^2\geq ID\geq\sum_i x_i^2 \geq\frac1n\left(\sum_i x_i\right)^2. \tag{3} \] The last inequality is Cauchy--Schwarz. Taking nonnegative square roots proves $1$. In fact the intermediate bound $M\geq(\sum x_i^2)^{1/2}$ is stronger than the normalized-sum assertion.

Allowing arbitrary real entries.

If some entries are positive, restrict to the subsequence of positive entries. It has length $m\leq n$ and positive sum $S_+$. If the full sum $S$ is positive, then $3$ gives \[ M\geq S_+/\sqrt m\geq S/\sqrt n. \] If $S\leq0$, any positive singleton already proves $1$. Finally, if every entry is nonpositive, the largest entry is at least the average $S/n$, and $S/n\geq S/\sqrt n$. Taking that singleton proves $1$ in the remaining case. Thus no positivity assumption is silently imposed on the original sequence.

Sharpness of the asymptotic constant.

Let $n=k^2$, and consider the permutation formed from $k$ consecutive decreasing blocks, \[ (k,k-1,\ldots,1),\ (2k,2k-1,\ldots,k+1),\ \ldots, (k^2,k^2-1,\ldots,k^2-k+1). \] Call its entries $r_i$. An increasing subsequence takes at most one entry from each block, and a decreasing subsequence lies in one block. Hence every monotone subsequence has at most $k$ entries.

For any $\eta>0$, set $x_i=1+\eta r_i/k^2$. These are distinct positive reals, with exactly the displayed order relations and all lying in $(1,1+\eta]$. Therefore \[ M\leq k(1+\eta),\qquad \sum_i x_i\geq k^2, \qquad \frac{\sqrt n\,M}{\sum_i x_i}\leq1+\eta. \] Let $k\to\infty$ and $\eta\to0$. This rules out every universal asymptotic coefficient greater than one. Normalizing each example by its total sum also shows that the square-length bound $1/k$ is the infimum, even with distinct positive entries.

Attribution and assessment.

The weighted conclusion appears in work of Tidor, Wang, and Yang and is implicit in work of Wagner. The square-packing exposition above follows the argument attributed to Aristotle in the discussion, described there by llllvvuu; KoishiChan supplies an alternative blow-up proof, and Cambie supplies the near-equal block examples. Sources: https://www.erdosproblems.com/1026 and https://www.erdosproblems.com/forum/thread/1026, checked 24 September 2026.

The exact input recurrence, universal bound, square-length conjecture, and asymptotic constant are completely proved here. No external solution theorem or local Lean verification is assumed, and no new result is claimed.

Completion rate estimate: 100%.